Question

Answer all assigned questions and problems, and show all work.

1. Determine the pH of (a) a 0.40 MCH₃CO₂H solution, (b) a solution that is 0.40 MCH₃CO₂H and 0.20 MNaCH₃CO₂. (8 points)

(Reference: Chang 16.5)

2. Which of the following solutions can act as a buffer? (a) KCl/HCl, (b) KHSO₄/H₂SO_4, (c) KNO₂/HNO₂.(5 points)

(Reference: Chang 16.9)

3. Calculate the pH of the buffer system made up of 0.15 MNH₃/0.35 MNH₄Cl. (5 points)

(Reference: Chang 16.11)

4. The pH of a bicarbonate-carbonic acid buffer is 8.00. Calculate the ratio of the concentration of carbonic acid (H₂CO₃) to that of the bicarbonate ion (HCO₃^–). (5 points)

(Reference: Chang 16.13)

5. The pH of a sodium acetate–acetic acid buffer is 4.50. Calculate the ratio [CH₃COO^–]/[CH₃COOH].(5 points)

(Reference: Chang 16.15)

6. Calculate the pH of the 0.20 MNH₃/0.20 MNH₄Cl buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 MHCl to 65.0 mL of the buffer? (8 points)
   

(Reference: Chang 16.17)

7. A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid. (5 points)

(Reference: Chang 16.27)

8. In a titration experiment, 12.5 mL of 0.500 MH₂SO₄neutralize 50.0 mL of NaOH. What is the concentration of the NaOH solution? (5 points)

(Reference: Chang 16.29)

9. A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633 MNaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the K_aof the unknown acid? (10 points)

(Reference: Chang 16.31)

            

10. Calculate the pH at the equivalence point for the following titration: 0.20 MHCl versus 0.20 Mmethylamine (CH₃NH₃;K_b= 4.4 × 10^–4)(5 points)

            

(Reference: Chang 16.33)

11. A 25.0 mL solution of 0.100 MCH₃COOH is titrated with a 0.200 MKOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL, (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL. (25 points)

            (Reference: Chang 16.35)

12. Referring to the following table:

Indicator	Color in Acid	Color in Base	pH Range*
Thymol blue	Red	Yellow	1.2–2.8
Bromophenol blue	Yellow	Bluish purple	3.0–4.6
Methyl orange	Orange	Yellow	3.1–4.4
Methyl red	Red	Yellow	4.2–6.3
Chlorophenol blue	Yellow	Red	4.8–6.4
Bromothymol blue	Yellow	Blue	6.0–7.6
Cresol red	Yellow	Red	7.2–8.8
Phenolphthalein	Colorless	Reddish pink	8.3–10.0

*The pH range is defined as the range over which the indicator changes from the acid color to the base color.

Specify which indicator or indicators you would use for the following titrations:  (9 points)

1. HCOOH versus NaOH
2. HCl versus KOH
3. HNO₃versus CH₃NH₂

(Reference: Chang 16.43)

13. The pK_aof butyric acid (HBut) is 4.7. Calculate K_bfor the butyrate ion (But^–). (5 points)

(Reference

Answer 1

& Finst questions : Determination ut ut i - Here the given acid is acetic acid (CHg. CooH). It is a weak acid. For any weak acid the concentration of Hgot ion is - [H30 t] - [ ka x = (exka Again by definition - pH = -log [ttso t] = - 12 log Ka - loge a pH = 2 pka - 22 loge [ipka = -log ka] ka=ionisation constant of acid c = concentration of weak acid.

Now given that I the concentration c = 0.4 M And. pka of CH2 LOUH = 4.76 (data collected). Now from equation ③ we have - pH = 1 (4.76) – 12 log 0.4 = 2.38 – 0.2 = 2.58 Now, the pH of 0.4M CH3COH solution = 2.58 (Aus) - Acid => CH3 COOH (weak). Salt = CHg cogna So, it is a solution of weak acid and its salt, so it acts as a buffers solution Now, for buffer solution the expression of pt can be given as

[salt] pH = pka + log [ Acid the get of Now, Here given, TAcid) = CH2, Cro H = 0-4 M [salt] = [CH3COONa] = 0.2 m for acetic acid, - O pka = 4.76. Now, froom eqr 2 we can pH of the solution as - pH = 4.76 + log Ich y [CHg cook = 4.76 + logo = 4.76 + (-0.3) = 4.46 & pH of the given solution = 4.46 fans) [CH₂ Loona 0.2

(Reference: Change 16-5) A buffen solution is the mixture of solutions of weak acid and its salt. Now HU and Hq so both are strong acid. So both HU and Hq So do not form buffes solution, so options @ and are ruled out. Again, Nitrous acid HNO 2 Weak acid Potassium nitrite KNO2 Salt Mixture of HNO and KNO2 can act as a buffers So the connect option is - (c) KNO2/HNO2 (Reference change 16.9) A mixture of weak base and its salt can also act as a buffer I Solution.

for these type of Solution - [salt] рон ерко тоГоле) Kb = ionisation constant of weak base Again, pH = 14 -POH Now given - [Base] = [Nitg] = 0.15 m Too NHg act as a weak base] [salt] - [Nitqul] = 0.35 m a. From equation ④ - pOH = pkb + log - [Nitqu Log In Ho] lopky of NH3 -4.757 a pot = 4.75 + log 0.35 o 0.15 = 4.75 +0.37 = 5.12 from equation 2 pH = 14-5.12 = 8.88 Ans
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